Systems of units: SI, oilfield and metric oilfield

Quick refresher

Everything follows from a few exact definitions: 1 ft = 0.3048 m, 1 in = 2.54 cm, 1 lb = 0.45359237 kg, the US gallon = 231 cubic inches, the barrel = 42 gallons = 0.158987 m³, 1 atm = 101,325 Pa, 1 cP = 1 mPa·s. From these we get 1 psi = 6894.757 Pa and 1 D = 0.9869·10⁻¹² m².
In SI the formulas contain no numerical constants. The numbers 1.127·10⁻³, 141.2, 0.0853, 0.536 and 2.637·10⁻⁴ are unit conversions, and each is valid only for its own set of units.
We always state the reference conditions with a volume of gas. A cubic metre at 20 °C holds 1.5% less gas than a cubic metre at 60 °F, and 6.8% less than one at 0 °C. In Russia and Kazakhstan we refer gas to 20 °C and 101.325 kPa. In Kazakhstan, oil and condensate reserves are accounted for by mass.
We put absolute pressure into the gas laws and into PVT calculations (PVT: how the volume of a fluid changes with pressure and temperature). Gauge pressure is too low by the atmospheric pressure. The lower the pressure, the larger the error: 17% at 0.5 MPa and 0.4% at 25 MPa.
Conversion factors with 4–5 significant digits give an error of hundredths of a per cent. That is less than any measurement error. The danger is not in the digits but in a mixed-up basis: reference conditions, gauge and absolute pressure, reservoir and standard volumes, natural and decimal logarithm, thousand and million.

Key relationships

1\ \text{bbl} = 42\ \text{gal} = 0.158987\ \text{m}^3 = 5.6146\ \text{ft}^3

the barrel is defined exactly through the gallon, and the gallon through the inch

1\ \text{psi} = 6894.757\ \text{Pa},\qquad 1\ \text{MPa} = 145.04\ \text{psi}

pressure

p_{\text{abs}} = p_{\text{gauge}} + 0.101325\ \text{MPa}

absolute pressure from a gauge reading

V_2 = V_1\,\frac{T_2}{T_1}\,\frac{p_1}{p_2}

the same amount of gas at different reference conditions; temperatures and pressures are absolute

{}^{\circ}\mathrm{API} = \frac{141.5}{\gamma_o} - 131.5

$\gamma_o$ — specific gravity of the oil relative to water at 60 °F

Common pitfalls

Putting a pressure gauge reading (gauge pressure, psig) into the equation of state of a gas (the equation that links pressure, volume and temperature).
Adding gas volumes that are referred to different conditions. Or converting the solution gas–oil ratio with the factor 5.615 when the cubic metres of gas are referred to 20 °C.
Reading Mscf as a million cubic feet. Here M is a thousand, and a million is MM.
Taking the constant 162.6 (for the decimal logarithm) for a formula with the natural logarithm. Or using a constant worked out for one set of units with data in another.
Putting a temperature in degrees Celsius or Fahrenheit where an absolute temperature is needed.

After this lesson you can

name the units of the main reservoir engineering quantities in SI, oilfield and metric oilfield units
work out conversion factors from the exact definitions of the foot, inch, pound, gallon, atmosphere and darcy, and decide how many significant digits to keep
state the reference conditions (the temperature and pressure to which a gas volume is stated) for every volume of gas, and convert volumes and the solution gas–oil ratio (the gas dissolved in the oil) between 20 °C, 15 °C, 60 °F and 0 °C
convert gauge pressure to absolute pressure, and convert temperatures between the Celsius, Kelvin, Fahrenheit and Rankine scales (Rankine is the absolute scale that uses Fahrenheit degrees)
convert oil density to specific gravity (density relative to water), degrees API (an oilfield scale for oil density) and barrels per tonne
work out the constant of a formula for the units you need, and check a formula by its dimensions

Three systems of units

SI is the basis of physics calculations: metre, kilogram, second, pascal, kelvin. In SI, Darcy's law (the law of flow through rock), the equation of state of a gas and the diffusivity equation (the equation for how a pressure change spreads through the reservoir) are written without numerical coefficients. This is why we derive formulas in SI. For the oilfield SI has one drawback: the numbers have an awkward size. The permeability of a good reservoir is about $10^{-13}$ m², and the flow rate of a well is a few thousandths of a cubic metre per second.

Oilfield units (field units) grew up in the American oil industry: feet, barrels per day, pounds-force per square inch (psi), centipoise, millidarcies and degrees Fahrenheit. Most of the world's literature and many programs are written in them, and international companies report in them. Metric oilfield units are metres, cubic metres per day, megapascals, mPa·s, millidarcies and degrees Celsius. They are familiar in Kazakhstan and Russia, and the lessons of the library are written in them. Both oilfield systems give numbers of a convenient size, but their formulas contain constants.

The main quantities in the three systems of units

Quantity: Length, thickness
SI: m
Oilfield: ft
Metric oilfield: m

Quantity: Pressure
SI: Pa
Oilfield: psi (psia, psig)
Metric oilfield: MPa

Quantity: Liquid rate
SI: m³/s
Oilfield: bbl/d (STB/d)
Metric oilfield: m³/d

Quantity: Gas rate
SI: m³/s
Oilfield: Mscf/d
Metric oilfield: thousand m³/d

Quantity: Permeability
SI: m²
Oilfield: mD
Metric oilfield: mD; µm²

Quantity: Viscosity
SI: Pa·s
Oilfield: cP
Metric oilfield: mPa·s

Quantity: Compressibility
SI: 1/Pa
Oilfield: 1/psi
Metric oilfield: 1/MPa

Quantity: Temperature
SI: K
Oilfield: °F; °R
Metric oilfield: °C; K

Quantity: Density
SI: kg/m³
Oilfield: lb/ft³, °API, ppg
Metric oilfield: kg/m³, g/cm³

Quantity: Time
SI: s
Oilfield: h, d
Metric oilfield: h, d

Quantity	SI	Oilfield	Metric oilfield
Length, thickness	m	ft	m
Pressure	Pa	psi (psia, psig)	MPa
Liquid rate	m³/s	bbl/d (STB/d)	m³/d
Gas rate	m³/s	Mscf/d	thousand m³/d
Permeability	m²	mD	mD; µm²
Viscosity	Pa·s	cP	mPa·s
Compressibility	1/Pa	1/psi	1/MPa
Temperature	K	°F; °R	°C; K
Density	kg/m³	lb/ft³, °API, ppg	kg/m³, g/cm³
Time	s	h, d	h, d

psia is absolute pressure and psig is gauge pressure, both in psi. STB is the stock-tank barrel (a barrel of oil at standard conditions), Mscf is a thousand standard cubic feet, and ppg is pounds per gallon (the density of drilling mud). In well tests we give time in hours, and in production in days. 1 mD ≈ 10⁻³ µm² (more precisely 0.987·10⁻³).

In Kazakhstan the Law “On Ensuring the Uniformity of Measurements” permits the use of the units of the International System. It permits other units, outside the system, by decision of the authorised body. In practice, reports and project documents use metres, megapascals, cubic metres and tonnes. American oilfield units appear in foreign literature, in programs and in the reports of international operators. In older Russian-language reports, pressure may be written in kgf/cm² or atmospheres (units of old reports).

Exact definitions and conversion factors

We do not have to memorise conversion factors or look them up in tables. Almost all of them follow from a few definitions that were fixed exactly. In 1959 the English-speaking countries agreed on an international yard and pound. Since then the foot is exactly 0.3048 m, the inch is exactly 2.54 cm and the pound is exactly 0.45359237 kg. By definition the US gallon equals 231 cubic inches, and the oil barrel (bbl) equals 42 gallons.

Exact definitions that the factors follow from

Unit: foot (ft)
Definition: international foot
In SI units: 0.3048 m exactly

Unit: inch (in)
Definition: 1/12 of a foot
In SI units: 0.0254 m exactly

Unit: pound (lb)
Definition: international pound
In SI units: 0.45359237 kg exactly

Unit: pound-force (lbf)
Definition: the weight of one pound under standard gravity (9.80665 m/s²)
In SI units: 4.448222 N

Unit: psi
Definition: pound-force per square inch
In SI units: 6894.757 Pa

Unit: US gallon (gal)
Definition: 231 cubic inches
In SI units: 3.785411784 L exactly

Unit: barrel (bbl)
Definition: 42 US gallons
In SI units: 0.158987 m³ — exactly 0.158987294928

Unit: atmosphere (atm)
Definition: standard atmosphere
In SI units: 101,325 Pa exactly

Unit: centipoise (cP)
Definition: 0.01 poise
In SI units: 1 mPa·s exactly

Unit: darcy (D)
Definition: through Darcy's law, see below
In SI units: 0.9869233·10⁻¹² m²

Unit	Definition	In SI units
foot (ft)	international foot	0.3048 m exactly
inch (in)	1/12 of a foot	0.0254 m exactly
pound (lb)	international pound	0.45359237 kg exactly
pound-force (lbf)	the weight of one pound under standard gravity (9.80665 m/s²)	4.448222 N
psi	pound-force per square inch	6894.757 Pa
US gallon (gal)	231 cubic inches	3.785411784 L exactly
barrel (bbl)	42 US gallons	0.158987 m³ — exactly 0.158987294928
atmosphere (atm)	standard atmosphere	101,325 Pa exactly
centipoise (cP)	0.01 poise	1 mPa·s exactly
darcy (D)	through Darcy's law, see below	0.9869233·10⁻¹² m²

The other factors follow from these definitions. A cubic foot is $0.3048^3 = 0.0283168$ m³, and a barrel contains $0.158987/0.0283168 = 5.6146$ cubic feet. A pound-force is the weight of a pound, and psi is that force on one square inch, an area of $0.0254^2$ m². The darcy is defined through Darcy's law itself. A rock of 1 D passes 1 cm³/s of a liquid of viscosity 1 cP through a sample with a cross-section of 1 cm² and a length of 1 cm, at a pressure difference of 1 atm (units of permeability). The advanced section shows how to convert these units to SI.

The conversion factors needed most often

Quantity: Length
Oilfield unit: 1 ft
In metric units: 0.3048 m
Reverse conversion: 1 m = 3.2808 ft

Quantity: Volume
Oilfield unit: 1 bbl
In metric units: 0.158987 m³
Reverse conversion: 1 m³ = 6.2898 bbl

Quantity: Volume
Oilfield unit: 1 ft³
In metric units: 0.0283168 m³
Reverse conversion: 1 m³ = 35.3147 ft³

Quantity: Pressure
Oilfield unit: 1 psi
In metric units: 6.894757 kPa
Reverse conversion: 1 MPa = 145.04 psi

Quantity: Pressure gradient
Oilfield unit: 1 psi/ft
In metric units: 22.621 kPa/m
Reverse conversion: 1 MPa/m = 44.21 psi/ft

Quantity: Permeability
Oilfield unit: 1 mD
In metric units: 0.9869·10⁻¹⁵ m²
Reverse conversion: 1 µm² = 1013 mD

Quantity: Viscosity
Oilfield unit: 1 cP
In metric units: 1 mPa·s
Reverse conversion: —

Quantity: Compressibility
Oilfield unit: 1 psi⁻¹
In metric units: 145.04 MPa⁻¹
Reverse conversion: 1 MPa⁻¹ = 6.895·10⁻³ psi⁻¹

Quantity: Density
Oilfield unit: 1 lb/ft³
In metric units: 16.018 kg/m³
Reverse conversion: 1 g/cm³ = 62.43 lb/ft³ = 8.345 ppg

Quantity: Gas rate
Oilfield unit: 1 Mscf/d
In metric units: 28.317 m³/d at the same reference conditions
Reverse conversion: 1000 m³/d = 35.31 Mscf/d

Quantity	Oilfield unit	In metric units	Reverse conversion
Length	1 ft	0.3048 m	1 m = 3.2808 ft
Volume	1 bbl	0.158987 m³	1 m³ = 6.2898 bbl
Volume	1 ft³	0.0283168 m³	1 m³ = 35.3147 ft³
Pressure	1 psi	6.894757 kPa	1 MPa = 145.04 psi
Pressure gradient	1 psi/ft	22.621 kPa/m	1 MPa/m = 44.21 psi/ft
Permeability	1 mD	0.9869·10⁻¹⁵ m²	1 µm² = 1013 mD
Viscosity	1 cP	1 mPa·s	—
Compressibility	1 psi⁻¹	145.04 MPa⁻¹	1 MPa⁻¹ = 6.895·10⁻³ psi⁻¹
Density	1 lb/ft³	16.018 kg/m³	1 g/cm³ = 62.43 lb/ft³ = 8.345 ppg
Gas rate	1 Mscf/d	28.317 m³/d at the same reference conditions	1000 m³/d = 35.31 Mscf/d

We rounded the factors to 4–7 significant digits. The exact values follow from the definitions above. The conversion “a thousand cubic feet is 28.317 m³” holds only for identical reference conditions. The section below covers them.

Worked exampleWell data sheet: from oilfield to metric units

Given (assumed data)

Net pay thickness $h$ (the thickness that can produce): 46 ft
Reservoir pressure $p$: 3480 psia
Permeability; oil viscosity: 85 mD; 1.9 cP
Oil rate $q$: 1250 STB/d
Total compressibility $c_t$: $12\cdot10^{-6}$ psi⁻¹
Reservoir temperature: 190 °F

Solution

Thickness: $46\cdot0.3048 = 14.02$ m.
Pressure: $3480\cdot0.006894757 = 23.99$ MPa — absolute pressure, as psia is.
Permeability and viscosity do not change: the millidarcy is common to both systems, and 1 cP = 1 mPa·s.
Rate: $1250\cdot0.158987 = 198.7$ m³/d at standard conditions.
Compressibility: there are 145.04 psi in a megapascal, so $c_t = 12\cdot10^{-6}\cdot145.04 = 1.740\cdot10^{-3}$ MPa⁻¹ — we multiply, not divide.
Temperature: $(190 - 32)/1.8 = 87.8$ °C, or 360.9 K.

Answer: In metric oilfield units:

h

= 14.02 m,

p

= 23.99 MPa,

k

= 85 mD,

\mu

= 1.9 mPa·s,

q

= 198.7 m³/d,

c_t

= 1.740·10⁻³ MPa⁻¹, temperature 87.8 °C.

What it means. We check the direction of a conversion with common sense. Compressibility is the change in volume per unit of pressure. A megapascal is 145 times larger than a psi, so per megapascal the volume changes 145 times more, and the number in 1/MPa is larger. The most frequent error in such data sheets is pressure: we must first convert a reading in psig to absolute pressure. The data in this example are assumed for illustration.

Pressure and temperature: from which zero they are measured

We measure pressure and temperature on scales with different zero points, and this is the main source of errors when we convert them. Absolute pressure is measured from vacuum, and gauge pressure from atmospheric pressure. A pressure gauge at the wellhead or on a separator (a vessel that splits the well stream into oil, gas and water) shows how much the pressure exceeds atmospheric. In oilfield units the difference is written into the unit itself: psia (absolute) and psig (gauge). In metric units we state it in words: “MPa, absolute” or “MPa gauge”.

Absolute pressure from a gauge reading

p_{\text{abs}} = p_{\text{gauge}} + p_{\text{atm}}

where:

$p_{\text{abs}}$ absolute pressure — measured from vacuum, MPa, psia
$p_{\text{gauge}}$ gauge pressure — the excess over atmospheric pressure, MPa, psig
$p_{\text{atm}}$ atmospheric pressure: the standard value is 0.101325 MPa = 14.696 psi = 1.01325 bar, MPa, psi

Units: any pressure units, the same for all three quantities

Holds when

For the conversion we usually take the standard atmospheric pressure. The actual pressure depends on altitude and weather. At 1000 m above sea level it is about 0.09 MPa.
Pressure differences do not depend on the choice of zero. Examples are drawdown (the pressure drop that pushes fluid into the well), the pressure drop across a choke (a restriction that controls the flow) and a pressure gradient.

The gas laws, the formation volume factors and the fluid properties depend on absolute pressure. In drilling and at the wellhead, people more often use gauge pressure, because the atmosphere presses on all sides and its effect cancels out (pore and overburden pressure).

CalculatedThe error from putting gauge pressure in place of absolute pressure. The pressure, and the amount of gas calculated from it, come out too low by this many per cent. In a separator at 0.5 MPa the error is 17%, at the wellhead at 5 MPa it is 2%, and in the reservoir at 25 MPa it is 0.4%.

Temperature works in the same way. The Celsius and Fahrenheit scales start from chosen points: ice melts at 0 °C and at 32 °F. The Kelvin and Rankine scales start from absolute zero (the lowest possible temperature). A degree Celsius is the same size as a kelvin. A degree Fahrenheit is the same size as a degree Rankine, and it is 1.8 times smaller than a kelvin. We put only absolute temperature into the gas laws.

Temperature scales

Point: Absolute zero
°C: −273.15
K: 0
°F: −459.67
°R: 0

Point: Melting of ice
°C: 0
K: 273.15
°F: 32
°R: 491.67

Point: US standard temperature
°C: 15.56
K: 288.71
°F: 60
°R: 519.67

Point: Standard temperature in Russia and Kazakhstan
°C: 20
K: 293.15
°F: 68
°R: 527.67

Point: Boiling of water at 1 atm
°C: 100
K: 373.15
°F: 212
°R: 671.67

Point	°C	K	°F	°R
Absolute zero	−273.15	0	−459.67	0
Melting of ice	0	273.15	32	491.67
US standard temperature	15.56	288.71	60	519.67
Standard temperature in Russia and Kazakhstan	20	293.15	68	527.67
Boiling of water at 1 atm	100	373.15	212	671.67

Temperature conversion

T\,[\text{K}] = t\,[{}^\circ\text{C}] + 273.15,\qquad t\,[{}^\circ\text{F}] = 1.8\,t\,[{}^\circ\text{C}] + 32,\qquad T\,[{}^\circ\text{R}] = t\,[{}^\circ\text{F}] + 459.67 = 1.8\,T\,[\text{K}]

where:

$T$ absolute temperature — in kelvins or degrees Rankine, K, °R
$t$ temperature on the Celsius or Fahrenheit scale, °C, °F

Units: given in square brackets

Holds when

We convert temperature differences without the shift (the added constant): 1 K = 1 °C, 1 °R = 1 °F, 1 K = 1.8 °R.
°R in oilfield formulas is the degree Rankine. In Russian physics reference books it is written °Ra, because there °R is the obsolete degree Réaumur. In older Russian-language literature the name Rankine is also transliterated as “Renkin”.
A geothermal gradient (how fast the temperature rises with depth) of 1 °F per 100 feet equals 1.823 °C per 100 m.

The constants 273.15 and 459.67 move the zero. The factor 1.8 changes the size of the degree. So converting temperatures needs both steps, while converting temperature differences and gradients needs only the second step.

Worked exampleWellhead and separator pressure, wellhead temperature

Given (assumed data)

Wellhead pressure gauge: 1840 psig
Wellhead temperature: 131 °F
Separator pressure gauge: 65 psig
Atmospheric pressure: 14.696 psi (0.101325 MPa)

Solution

Wellhead: $1840 + 14.696 = 1854.7$ psia, or $1854.7\cdot0.006894757 = 12.79$ MPa.
If we put 1840 into a gas law instead of 1854.7, the pressure and the amount of gas come out too low by $14.696/1854.7 = 0.79$ %.
Separator: $65 + 14.696 = 79.7$ psia $= 0.55$ MPa. The error from using the gauge pressure is $14.696/79.7 = 18.4$ %.
Temperature: $(131 - 32)/1.8 = 55.0$ °C; absolute — $55.0 + 273.15 = 328.15$ K, or $131 + 459.67 = 590.67$ °R. Check: $1.8\cdot328.15 = 590.67$ .

Answer: At the wellhead the absolute pressure is 12.79 MPa (putting in the gauge reading would give an error of 0.79%). In the separator it is 0.55 MPa (error 18.4%). The wellhead temperature is 328.15 K, or 590.67 °R.

What it means. The lower the pressure, the more the atmospheric addition matters. At the wellhead we can almost neglect it, while in the separator it changes the result by nearly a fifth. For pressure differences, such as drawdown or the drop across a choke, the zero does not matter. The data in this example are assumed for illustration.

Standard and normal conditions

The volume of a gas depends strongly on pressure and temperature. So at the surface we refer it to stated standard conditions. A cubic metre of gas at standard conditions is in effect a measure of the amount of substance, a definite number of moles (a mole is a fixed number of molecules) (molar volume). But there are several “standard” conditions, and the same gas takes up a different volume at each.

Reference conditions for gas volumes

Where and by whom adopted: Russia, Kazakhstan: GOST 2939-63, dry gas
Temperature: 20 °C
Pressure (absolute): 101.325 kPa (760 mmHg)
Molar volume, m³/kmol: 24.06

Where and by whom adopted: Kazakhstan: gas reserves under the Unified Rules
Temperature: 20 °C
Pressure (absolute): 0.1 MPa
Molar volume, m³/kmol: —

Where and by whom adopted: International standard ISO 13443 for natural gas
Temperature: 15 °C
Pressure (absolute): 101.325 kPa
Molar volume, m³/kmol: 23.64

Where and by whom adopted: US oilfield units
Temperature: 60 °F (15.56 °C)
Pressure (absolute): 14.696 psia (101.325 kPa)
Molar volume, m³/kmol: 23.69

Where and by whom adopted: Gas statistics and contracts in the US
Temperature: 60 °F
Pressure (absolute): often 14.73 psia
Molar volume, m³/kmol: 23.64

Where and by whom adopted: Normal conditions (physics, chemistry, the “normal cubic metre”)
Temperature: 0 °C
Pressure (absolute): 101.325 kPa
Molar volume, m³/kmol: 22.41

Where and by whom adopted	Temperature	Pressure (absolute)	Molar volume, m³/kmol
Russia, Kazakhstan: GOST 2939-63, dry gas	20 °C	101.325 kPa (760 mmHg)	24.06
Kazakhstan: gas reserves under the Unified Rules	20 °C	0.1 MPa	—
International standard ISO 13443 for natural gas	15 °C	101.325 kPa	23.64
US oilfield units	60 °F (15.56 °C)	14.696 psia (101.325 kPa)	23.69
Gas statistics and contracts in the US	60 °F	often 14.73 psia	23.64
Normal conditions (physics, chemistry, the “normal cubic metre”)	0 °C	101.325 kPa	22.41

The molar volumes are for an ideal gas (a gas whose molecules do not attract or repel each other), $V_m = RT/p$ . Natural gas at these conditions differs from an ideal gas by a fraction of a per cent. GOST 2939-63 “Gases. Conditions for determination of volume” (GOST is the state standard of the USSR and later of Russia) has been in force since 1964. Kazakhstan's normative documents on gas accounting also refer to it.

For oil, the difference between the conditions is small: between 15 and 20 °C the volume of stock-tank oil changes by about 0.3–0.4%. For gas it is large. A cubic metre at 20 °C holds 1.5% less gas than a cubic metre at 60 °F, and 6.8% less than a normal cubic metre at 0 °C. The same gas at 20 °C takes up a volume 7.3% greater than at 0 °C. For a field with gas reserves of tens of billions of cubic metres, that is hundreds of millions of cubic metres appearing from nothing. So there is one rule: give every volume of gas with its reference conditions, and bring volumes to the same conditions before adding or comparing them.

Conversion of a gas volume to other reference conditions

V_2 = V_1\,\frac{T_2}{T_1}\,\frac{p_1}{p_2}

where:

$V_1,\ V_2$ volume of the same amount of gas at conditions 1 and 2, m³, ft³
$T_1,\ T_2$ absolute temperatures of the reference conditions, K or °R
$p_1,\ p_2$ absolute pressures of the reference conditions, the same units

Units: temperatures absolute, pressures absolute and in the same units

Holds when

The amount of gas is the same. At reference conditions we treat the gas as ideal: the gas deviation factor (Z, which shows how far a real gas departs from an ideal one) is close to one, and its small difference almost cancels.
The moisture content is the same: under GOST 2939-63 the gas is dry, and water vapour is not included in the volume.

The same gas, a different number of cubic metres. When we convert the solution gas–oil ratio from m³/m³ to scf/STB, there is one more step: a cubic metre of oil must be converted into barrels.

Solution gas–oil ratio and producing gas–oil ratio: m³/m³ and scf/STB

R\,[\text{scf/STB}] = 5.6146\,R\,[\text{m}^3/\text{m}^3]\qquad\text{or}\qquad R\,[\text{scf/STB}] = 5.529\,R\,[\text{m}^3/\text{m}^3]

where:

$R$ solution gas–oil ratio or producing gas–oil ratio — the volume of gas per volume of oil at standard conditions, m³/m³; scf/STB

Units: gas — standard cubic metres or cubic feet, oil — cubic metres or barrels at standard conditions

Holds when

We use 5.6146 when the gas on both sides is referred to the same conditions. It is simply the number of cubic feet in a barrel.
We use 5.529 when the cubic metres of gas are referred to 20 °C and 101.325 kPa, and the cubic feet to 60 °F and 14.696 psia.
The volume of oil between 20 °C and 60 °F changes by about 0.35%. We usually neglect this.

The lessons of the library use both factors and state the conditions each time. They use 5.615 for identical conditions (Black oil model: solution gas–oil ratio and formation volume factors) and 5.53 for a conversion from 20 °C to 60 °F (PVT correlations for oil and gas condensate).

In other unit systems (1)

Where 5.529 comes from

5.529 = 5.61458\cdot\frac{288.706}{293.15}

the number of cubic feet in a barrel, multiplied by the ratio of the absolute temperatures of 60 °F and 20 °C. The two pressures, 14.696 psia and 101.325 kPa, are the same.

Gas measured at 20 °C shrinks when it is cooled to 60 °F (15.56 °C): the same gas takes up a volume 1.5% smaller. So the factor is smaller than the purely geometric 5.6146.

Worked exampleOne and the same gas in different cubic metres

Given (assumed data)

Separator gas for a month: 2.50 million m³ at 20 °C and 101.325 kPa (per GOST 2939-63)
Solution gas–oil ratio of the oil: 96 m³/m³ (gas at 20 °C)

Solution

To 15 °C (ISO 13443): $2.50\cdot288.15/293.15 = 2.457$ million m³.
To normal conditions, 0 °C: $2.50\cdot273.15/293.15 = 2.329$ million m³.
In standard cubic feet at 60 °F (288.71 K) and 14.696 psia — the same pressure as 101.325 kPa: in a cubic metre at 20 °C there are $35.3147\cdot288.71/293.15 = 34.78$ scf, and $2.50\cdot34.78 = 86.95$ million scf (MMscf).
With a reference pressure of 14.73 psia: $86.95\cdot14.696/14.73 = 86.75$ MMscf.
If we take the reference pressure of 0.1 MPa literally: $2.50\cdot101.325/100 = 2.533$ million m³.
Solution gas–oil ratio: $96\cdot5.529 = 531$ scf/STB; with the factor 5.6146 it would come to 539 scf/STB — 1.5% more.

Answer: The same 2.50 million m³ at 20 °C is 2.457 million m³ at 15 °C, 2.329 million normal cubic metres and 86.95 MMscf at 60 °F and 14.696 psia; a solution gas–oil ratio of 96 m³/m³ is 531 scf/STB.

What it means. The amount of gas is the same in every line. Only the “cubic metre” that we measure it in changes. Differences of 1.5–7% are larger than the error of a good gas meter. So a gas volume without reference conditions has no meaning, and we cannot add totals from different sources without converting them to the same conditions. We treated the gas as ideal at the reference conditions. The compressibility correction at atmospheric pressure is a fraction of a per cent, and it almost cancels. The data in this example are assumed for illustration.

Density: kg/m³, specific gravity and degrees API

In Russian-language practice, we give the density of stock-tank oil (oil that has lost its dissolved gas at the surface) in kg/m³ or g/cm³ at 20 °C. In international practice, it is given as the specific gravity $\gamma_o$ (the density relative to water) at 60 °F, or in degrees API (the API scale). The density of water at 60 °F is 999.0 kg/m³, so $\gamma_o$ almost equals the density in g/cm³. But we must refer both densities to the same temperature, because oil expands by about 0.08% when heated by 1 °C.

Specific gravity of oil from its density at 20 °C

\rho_{60} = \rho_{20}\left[1 + \beta\,(20 - 15.56)\right],\qquad \gamma_o = \frac{\rho_{60}}{999.0}

where:

$\rho_{20},\ \rho_{60}$ density of stock-tank oil at 20 °C and at 60 °F (15.56 °C), kg/m³
$\beta$ volumetric thermal expansion coefficient of the oil, 1/°C
$\gamma_o$ specific gravity of the oil relative to water at 60 °F, —

Units: kg/m³; 999.0 kg/m³ is the density of water at 60 °F

Holds when

We assume linear expansion over a narrow temperature range. $\beta \approx 8\cdot10^{-4}$ 1/°C is a typical value for medium oils, and the lessons of the library use it. For commercial transactions, the conversion is made with standard tables.
The density is that of stock-tank oil at atmospheric pressure.

Converting from 20 °C to 15.56 °C adds about 0.36% to the density and lowers the degrees API by about 0.6.

The number of barrels in a tonne depends on density. A tonne is $1/\rho$ cubic metres, and a cubic metre is 6.2898 barrels. Strictly, we take the density at the temperature to which the volume is referred. For a barrel this is 60 °F.

Barrels per tonne of oil

n_{\text{bbl/t}} = \frac{6.2898}{\rho\,[\text{t/m}^3]}

where:

$n_{\text{bbl/t}}$ number of barrels in a tonne of oil, bbl/t
$\rho$ density of the oil at the temperature to which the barrel volume is referred (60 °F), t/m³

Units: 6.2898 is the number of barrels in a cubic metre

Holds when

The oil is stock-tank oil. The density and the volume are at the same temperature.

At 0.86 t/m³ a tonne holds 7.31 barrels, at 0.80 it holds 7.86, and at 0.92 it holds 6.84. A light oil gives more barrels per tonne.

We give the specific gravity of a gas relative to air: $\gamma_g = M/28.97$ , where $M$ is the molar mass of the gas, the mass of one mole (gas specific gravity, Ideal and real gas: the gas deviation factor (Z)). In oilfield units, the density of drilling mud (the fluid pumped down the well while it is drilled) is given in pounds per gallon (ppg): 1 g/cm³ = 8.345 ppg (Drilling fluids: functions and properties).

Worked exampleOil: specific gravity, degrees API and barrels per tonne

Given (assumed data)

Density of stock-tank oil at 20 °C: 828 kg/m³
Thermal expansion coefficient of the oil: $8\cdot10^{-4}$ 1/°C (as in the lesson PVT correlations for oil and gas condensate)
Density of water at 60 °F: 999.0 kg/m³
Oil production: 1500 t/d

Solution

Density at 60 °F = 15.56 °C: $828\cdot(1 + 8\cdot10^{-4}\cdot4.44) = 830.9$ kg/m³ — cooled oil is denser.
Specific gravity: $\gamma_o = 830.9/999.0 = 0.832$ ; degrees API: $141.5/0.832 - 131.5 = 38.6$ .
Barrels per tonne: $6.2898/0.8309 = 7.57$ .
If we use the density at 20 °C, the result is $6.2898/0.828 = 7.60$ — larger by $(830.9/828 - 1)\cdot100 = 0.35$ %. But these would be barrels at 20 °C, not at 60 °F.
Daily production: $1500\cdot7.57 = 11{,}355$ bbl/d.

Answer: The oil is 38.6 °API (specific gravity 0.832), with 7.57 barrels per tonne; 1500 t/d is about 11,355 bbl/d.

What it means. The temperature correction changes the density by only a third of a per cent, but it moves the degrees API by a few tenths and the number of barrels by 0.35%. When oil is accounted for in a sale, such a difference is money. So we convert densities and volumes with standard tables, not with an average correction as here. The data in this example are assumed for illustration.

Where the constants in formulas come from

Darcy's law in SI contains no numbers: $q = kA\Delta p/(\mu L)$ . Suppose we want to put oilfield quantities into it — millidarcies, feet, psi, centipoise — and get barrels per day. Then we convert each quantity to SI, and we convert the result back. All the factors collect into one number, the constant of the formula.

Worked exampleWorking out the constant of Darcy's law in oilfield units

Given (assumed data)

Darcy's law in SI: $q = kA\Delta p/(\mu L)$ , flow rate in m³/s
Oilfield units: $k$ in mD, $A$ in ft², $\Delta p$ in psi, $\mu$ in cP, $L$ in ft; $q$ in bbl/d
Factors: 1 mD = $0.9869233\cdot10^{-15}$ m²; 1 ft = 0.3048 m; 1 psi = 6894.757 Pa; 1 cP = $10^{-3}$ Pa·s; 1 bbl = 0.158987 m³; 86,400 s in a day

Solution

Area: a square foot is $0.3048^2 = 0.09290304$ m². We multiply the numerator $kA\Delta p$ by $0.9869233\cdot10^{-15}$ , by 0.09290304 and by 6894.757 — altogether by $6.3217\cdot10^{-13}$ .
We multiply the denominator $\mu L$ by $10^{-3}\cdot0.3048 = 3.048\cdot10^{-4}$ . The division gives $2.0740\cdot10^{-9}$ . With this factor the flow rate comes out in cubic metres per second.
Convert m³/s to barrels per day: multiply by $86{,}400/0.158987 = 543{,}441$ .
The constant: $2.0740\cdot10^{-9}\cdot543{,}441 = 1.1271\cdot10^{-3}$ .
In the radial flow formula (flow towards a well from all sides) the flow area is the side surface of a cylinder, $2\pi rh$ . So we multiply the constant by $2\pi$ and get $7.082\cdot10^{-3}$ . Its reciprocal is 141.2.

Answer: The constant of Darcy's law in oilfield units is

1.1271\cdot10^{-3}

; in the radial flow formula it is

7.082\cdot10^{-3}

, with the reciprocal 141.2.

What it means. The constant packs the unit conversions of six quantities ( $k$ , $A$ , $\Delta p$ , $\mu$ , $L$ and $q$ ) into one number. It is valid only for the set of units that it was worked out for. The same route from the definitions gives the metric constant 0.0853 (mD, m², MPa, mPa·s, m and m³/d) and all the constants of the table below.

The metric constant is simpler. The lengths are already in metres. We multiply the permeability in mD by $0.9869\cdot10^{-15}$ , the pressure in MPa by $10^{6}$ , and the viscosity in mPa·s by $10^{-3}$ (it is in the denominator, so the constant is divided by $10^{-3}$ ). We multiply the flow rate in m³/s by the 86,400 seconds in a day. The result is 0.0853 (Darcy's law). The other constants of the library come out in the same way, and it is useful to know where each comes from:

Constants of formulas in the lessons of the library

Formula: Darcy's law, linear flow
Oilfield units: $1.127\cdot10^{-3}$
Metric oilfield units: 0.0853
Lesson: Rock permeability

Formula: Radial inflow (Dupuit equation)
Oilfield units: $7.08\cdot10^{-3}$ ; 141.2
Metric oilfield units: 0.536; 1.866
Lesson: Steady-state radial flow

Formula: Slope of the semilog straight line
Oilfield units: 162.6 (lg); 70.6 (ln)
Metric oilfield units: 2.149 (lg); 0.933 (ln)
Lesson: where the constants come from

Formula: Dimensionless time, $t$ in hours
Oilfield units: $2.637\cdot10^{-4}$
Metric oilfield units: $3.553\cdot10^{-3}$
Lesson: Diffusivity equation

Formula: Gas formation volume factor
Oilfield units: 0.02828 ft³/scf; 0.00504 rb/scf
Metric oilfield units: $3.456\cdot10^{-4}$ at 20 °C
Lesson: Gas formation volume factor, compressibility and viscosity

Formula: Hydrostatic pressure
Oilfield units: 0.052 psi/ft per ppg
Metric oilfield units: 0.00981 MPa/m per g/cm³
Lesson: Pore pressure and overburden pressure

Formula: Barrels in an acre-foot (the volume of one acre of area, one foot deep)
Oilfield units: 7758
Metric oilfield units: —
Lesson: Volumetric estimation of hydrocarbons in place

Formula	Oilfield units	Metric oilfield units	Lesson
Darcy's law, linear flow	$1.127\cdot10^{-3}$	0.0853	Rock permeability
Radial inflow (Dupuit equation)	$7.08\cdot10^{-3}$ ; 141.2	0.536; 1.866	Steady-state radial flow
Slope of the semilog straight line	162.6 (lg); 70.6 (ln)	2.149 (lg); 0.933 (ln)	where the constants come from
Dimensionless time, $t$ in hours	$2.637\cdot10^{-4}$	$3.553\cdot10^{-3}$	Diffusivity equation
Gas formation volume factor	0.02828 ft³/scf; 0.00504 rb/scf	$3.456\cdot10^{-4}$ at 20 °C	Gas formation volume factor, compressibility and viscosity
Hydrostatic pressure	0.052 psi/ft per ppg	0.00981 MPa/m per g/cm³	Pore pressure and overburden pressure
Barrels in an acre-foot (the volume of one acre of area, one foot deep)	7758	—	Volumetric estimation of hydrocarbons in place

We calculated all values from the exact definitions of this lesson, and they agree with those printed in the lessons. lg is the decimal logarithm and ln is the natural one. 0.052 is the rounded value of $12/231 = 0.05195$ . 0.433 psi/ft (more precisely 0.4335) is the pressure gradient of water with a density of 1 g/cm³.

Worked exampleOne well in three systems of units

Given (assumed data)

Permeability $k$ ; thickness $h$: 35 mD; 22 m
Oil viscosity $\mu$ ; formation volume factor $B$: 3.2 mPa·s; 1.08 m³/m³
Drawdown $p_e - p_w$: 2.6 MPa
Radii $r_e$ and $r_w$: 220 m and 0.08 m

Solution

$\ln(r_e/r_w) = \ln(220/0.08) = 7.919$ .
SI: $k = 35\cdot0.9869\cdot10^{-15} = 3.454\cdot10^{-14}$ m², $\Delta p = 2.6\cdot10^{6}$ Pa, $\mu = 3.2\cdot10^{-3}$ Pa·s. The flow rate in the reservoir is $2\pi kh\Delta p/(\mu\ln(r_e/r_w)) = 4.899\cdot10^{-4}$ m³/s. At standard conditions it is $4.899\cdot10^{-4}/1.08\cdot86{,}400 = 39.19$ m³/d.
Metric oilfield units: $q = \dfrac{0.536\cdot35\cdot22\cdot2.6}{3.2\cdot1.08\cdot7.919} = 39.21$ m³/d.
Oilfield units: $h = 22/0.3048 = 72.18$ ft, $\Delta p = 2.6\cdot145.04 = 377.1$ psi. The numerator $7.08\cdot10^{-3}\cdot35\cdot72.18\cdot377.1 = 6744.9$ , the denominator $3.2\cdot1.08\cdot7.919 = 27.368$ , and $q = 6744.9/27.368 = 246.5$ STB/d.
Back to cubic metres: $246.5\cdot0.158987 = 39.19$ m³/d.

Answer: All three calculations give the same result: 39.19, 39.21 and 39.19 m³/d (246.5 STB/d).

What it means. The difference of hundredths of a per cent comes from rounding the constants 0.536 and 7.08·10⁻³. A difference of several per cent would mean an error in the units. A check calculation in two systems like this is the simplest test of any program or table. The data in this example are assumed for illustration.

Checking dimensions

Any physically correct equation has the same dimension on its left and right sides (it is “homogeneous”). A quantity under a logarithm or in the exponent of an exponential has no dimension. This gives a quick check of a formula and of the steps we take to change it.

For example, Darcy's law gives the dimension of permeability: $[k] = [q][\mu][L]/([A][\Delta p])$ , that is m³/s · Pa·s · m / (m² · Pa) = m². Permeability is an area. This is why we express it in square metres or square micrometres.

How to check a formula with a constant

1
Find the units of the constant
A constant without stated units is useless: 141.2 works only with mD, feet, psi, cP and barrels per day.
2
Check the dimensions in SI
Leave out the constant and put in the dimensions of the quantities. The result must match the dimension of the left-hand side. The arguments of logarithms and exponentials must have no dimension.
3
Recalculate the constant
Convert all quantities to SI and collect the factors. The printed constant must come out.
4
Make a check calculation
Calculate one case in two systems of units. The results must agree to within the rounding of the constants.

Worked exampleDimensionless time in two systems of units

Given (assumed data)

Permeability $k$ ; time $t$: 35 mD; 10 h
Porosity $\phi$ ; viscosity $\mu$: 0.20; 3.2 mPa·s
Total compressibility $c_t$ ; wellbore radius $r_w$: $1.5\cdot10^{-3}$ MPa⁻¹; 0.08 m

Solution

Dimension in SI: $kt/(\phi\mu c_t r_w^2)$ is m²·s / (Pa·s · Pa⁻¹ · m²). This is a dimensionless quantity. So, with units other than SI, the factor in front of $kt$ is a pure unit conversion.
Metric oilfield units (mD, h, mPa·s, 1/MPa, m): the factor $0.9869\cdot10^{-15}\cdot3600\cdot10^{9} = 3.553\cdot10^{-3}$ — as in the lesson Diffusivity equation.
Oilfield units (mD, h, cP, 1/psi, ft): divide $0.9869\cdot10^{-15}\cdot3600\cdot6894.76$ by $10^{-3}\cdot0.3048^2$ and get $2.637\cdot10^{-4}$ .
The metric calculation: the numerator $3.553\cdot10^{-3}\cdot35\cdot10 = 1.2436$ , the denominator $0.20\cdot3.2\cdot1.5\cdot10^{-3}\cdot0.08^2 = 6.144\cdot10^{-6}$ , and $t_D = 2.024\cdot10^{5}$ .
The same case in oilfield units: $c_t = 1.5\cdot10^{-3}/145.04 = 1.0342\cdot10^{-5}$ psi⁻¹, $r_w = 0.08/0.3048 = 0.2625$ ft, and $t_D = 2.024\cdot10^{5}$ — the same number.

Answer: The factors

3.553\cdot10^{-3}

and

2.637\cdot10^{-4}

are the same unit conversion written for two systems. The dimensionless time is

2.024\cdot10^{5}

in both.

What it means. A dimensionless quantity does not depend on the system of units. This is the point of dimensionless variables in well testing (Line-source solution and the pressure function). If $t_D$ differs between the two systems, there is an error in one of the constants or in the conversion of the data. The data in this example are assumed for illustration.

How many digits to keep and where the real mistakes are made

We seldom know the input data of development better than to a few per cent. Permeability is known at best to 10–20%, thickness and viscosity to a few per cent, and the formation volume factor to about one per cent. So it is enough to keep conversion factors to 4–5 significant digits and the constants of formulas to 3–4. Their rounding gives an error of hundredths of a per cent. We need exact values when numbers are converted back and forth many times, or when we reconcile large cumulative volumes, because there the discrepancies add up. We round only the final result, to the accuracy of the input data.

The price of rounding and the price of a mixed-up basis

What was done: A barrel taken as 0.159 m³ instead of 0.158987
Error in the result: +0.008%

What was done: 1 MPa taken as 145 psi instead of 145.04
Error in the result: −0.03%

What was done: Darcy constant 0.0853 instead of 0.08527
Error in the result: +0.03%

What was done: 1 mD taken as $10^{-15}$ m² (that is, $10^{-3}$ µm²)
Error in the result: +1.3%

What was done: A gas volume at 20 °C compared with a volume at 60 °F without conversion
Error in the result: 1.5%

What was done: Reference pressure of 0.1 MPa instead of 101.325 kPa
Error in the result: 1.3%

What was done: Normal cubic metres (0 °C) taken for standard ones (20 °C)
Error in the result: −6.8%

What was done: Gauge pressure of 0.5 MPa instead of absolute
Error in the result: −17%

What was done: The constant for the decimal logarithm in a formula with the natural one
Error in the result: a factor of 2.303

What was done: Darcy instead of millidarcy; MMscf instead of Mscf
Error in the result: a factor of 1000

What was done: Reservoir volume instead of volume at standard conditions
Error in the result: a factor of $B$ (1.1–2 for oil)

What was done	Error in the result
A barrel taken as 0.159 m³ instead of 0.158987	+0.008%
1 MPa taken as 145 psi instead of 145.04	−0.03%
Darcy constant 0.0853 instead of 0.08527	+0.03%
1 mD taken as $10^{-15}$ m² (that is, $10^{-3}$ µm²)	+1.3%
A gas volume at 20 °C compared with a volume at 60 °F without conversion	1.5%
Reference pressure of 0.1 MPa instead of 101.325 kPa	1.3%
Normal cubic metres (0 °C) taken for standard ones (20 °C)	−6.8%
Gauge pressure of 0.5 MPa instead of absolute	−17%
The constant for the decimal logarithm in a formula with the natural one	a factor of 2.303
Darcy instead of millidarcy; MMscf instead of Mscf	a factor of 1000
Reservoir volume instead of volume at standard conditions	a factor of $B$ (1.1–2 for oil)

The upper rows show the rounding of factors. The errors are smaller than any measurement error. The lower rows show a mixed-up basis, with errors from a few per cent to a thousandfold.

Another frequent error is to mix up reservoir volumes and surface volumes. We write a barrel at reservoir conditions as rb (reservoir barrel) and a barrel at standard conditions as STB. The formation volume factor links them (Black oil model: solution gas–oil ratio and formation volume factors). A rate of “200 barrels” with no clarification can mean either, and the difference is a factor of 1.2–1.5 for ordinary oil.

Terms and symbols

SI: The International System of Units: metre, kilogram, second, pascal, kelvin. The formulas of physics are written in it without numerical constants.
Oilfield units: The American system of petroleum practice: feet, barrels, psi, centipoise, millidarcies, °F. Formulas in it contain conversion constants.
Metric oilfield units: Metres, cubic metres per day, megapascals, mPa·s, millidarcies, °C. They are familiar in Kazakhstan and Russia, and the lessons of the library use them.
psia, psig: Absolute and gauge pressure in pounds-force per square inch. The difference is the atmospheric pressure, 14.696 psi.
bbl, STB, rb: The barrel (42 US gallons = 0.158987 m³). STB is a barrel of oil at standard conditions, and rb is a barrel at reservoir conditions.
scf, Mscf, MMscf: The standard cubic foot of gas, and a thousand or a million such feet.
Standard and normal conditions: The pressure and temperature to which gas volumes are referred. Standard conditions are 20 °C and 101.325 kPa in Russia and Kazakhstan, 15 °C by ISO 13443, and 60 °F and 14.696 psia in US oilfield units. Normal conditions are 0 °C.
°API, $\gamma_o$: °API is an oilfield scale of oil density. $\gamma_o$ is the specific gravity of oil relative to water at 60 °F.
Formula constant: A number made up of unit conversions. It is valid only for the set of units for which it was worked out.

Common mistakes and misconceptions

Wrong

“1 m³/m³ is always 5.615 scf/STB.”

In fact

Only when the gas on both sides is referred to the same conditions. If the cubic metres of gas are referred to 20 °C and the feet to 60 °F, the factor is 5.529.

Wrong

“Let us put the gauge pressure into the equation of state.”

In fact

A pressure gauge shows gauge pressure, but the gas laws and PVT calculations need absolute pressure. We add the atmospheric pressure, 0.101 MPa (14.7 psi), to the reading.

Wrong

“Mscf is a million cubic feet.”

In fact

M is the Roman thousand: Mscf is a thousand standard cubic feet, and MMscf is a million.

Wrong

“The more digits in a factor, the more accurate the result.”

In fact

The input data limit the accuracy. Factors with 4–5 digits are enough. It is more important not to mix up the basis: reference conditions, absolute and gauge pressure, reservoir and standard volumes.

Wrong

“The constant 141.2 suits any units; it is a property of the formula.”

In fact

It is a unit conversion: 141.2 is for mD, feet, psi, cP and barrels per day. In metric oilfield units the same formula contains 1.866.

In depth: advanced detail

The darcy from its definition

The darcy is defined through Darcy's law itself, not through metres. So its value in square metres is not a round number. We put the definition into Darcy's law written in SI:

The darcy in SI units

1\ \text{D} = \frac{(10^{-6}\ \text{m}^3/\text{s})\cdot(10^{-3}\ \text{Pa·s})\cdot(10^{-2}\ \text{m})}{(10^{-4}\ \text{m}^2)\cdot(101{,}325\ \text{Pa})} = 0.9869233\cdot10^{-12}\ \text{m}^2

where:

$10^{-6}\ \text{m}^3/\text{s}$ flow rate of 1 cm³/s, m³/s
$10^{-3}\ \text{Pa·s}$ viscosity of 1 cP, Pa·s
$10^{-2}\ \text{m};\ 10^{-4}\ \text{m}^2$ length of 1 cm and cross-sectional area of 1 cm², m, m²
$101{,}325\ \text{Pa}$ pressure difference of 1 atm, Pa

Units: SI

Holds when

The definition of the darcy: 1 cm³/s of a liquid of viscosity 1 cP through a sample of cross-section 1 cm² and length 1 cm at a pressure difference of 1 atm.
The atmosphere here is the standard atmosphere, exactly 101,325 Pa. With the technical atmosphere (1 kgf/cm²) the number would be different.

In Russian-language practice, permeability is also written in µm². There 1 D ≈ 1 µm² and 1 mD ≈ 10⁻³ µm², with an error of 1.3%. Remember this rounding when you compare core, well log and well test results from different sources.

Units in old reports

Russian-language materials of past decades use some units that are now almost forgotten. We convert them from their exact definitions:

Old and rare units

Unit: technical atmosphere, kgf/cm² (at)
Definition: a force of 1 kgf per 1 cm²
In SI units: 98,066.5 Pa ≈ 0.0981 MPa

Unit: physical (standard) atmosphere, atm
Definition: a pressure of 760 mmHg
In SI units: 101,325 Pa

Unit: millimetre of mercury, mmHg
Definition: 1/760 atm
In SI units: 133.322 Pa

Unit: bar
Definition: $10^5$ Pa
In SI units: 0.1 MPa

Unit: poise, P; centistokes, cSt
Definition: units of dynamic and kinematic viscosity
In SI units: 1 P = 0.1 Pa·s; 1 cSt = $10^{-6}$ m²/s

Unit: dyne per centimetre, dyn/cm
Definition: interfacial tension
In SI units: 1 mN/m

Unit	Definition	In SI units
technical atmosphere, kgf/cm² (at)	a force of 1 kgf per 1 cm²	98,066.5 Pa ≈ 0.0981 MPa
physical (standard) atmosphere, atm	a pressure of 760 mmHg	101,325 Pa
millimetre of mercury, mmHg	1/760 atm	133.322 Pa
bar	$10^5$ Pa	0.1 MPa
poise, P; centistokes, cSt	units of dynamic and kinematic viscosity	1 P = 0.1 Pa·s; 1 cSt = $10^{-6}$ m²/s
dyne per centimetre, dyn/cm	interfacial tension	1 mN/m

1 atm = 1.0332 kgf/cm². The two “atmospheres” differ by 3.3%, so in an old report it matters which of them is meant. 1 atm = 14.696 psi.

Kinematic viscosity $\nu = \mu/\rho$ (m²/s, cSt) is not the same as dynamic viscosity. Capillary viscometers (instruments in which the fluid runs through a thin tube) measure exactly the kinematic viscosity, and we need the density to convert it to mPa·s (Oil viscosity and interfacial tension).

Connections to other topics

Rock permeability — Darcy's law and the units of permeability — where 0.0853 and 1.127·10⁻³ come from.
Ideal and real gas: the gas deviation factor (Z) — the equation of state, molar volume and standard conditions for gas.
Reservoir pressure and temperature — reservoir pressure and temperature, absolute and gauge pressure in field development.
Black oil model: solution gas–oil ratio and formation volume factors — reservoir and standard volumes, the solution gas–oil ratio in m³/m³ and scf/STB.
Composition of reservoir fluids — oil density and the API scale.
Pressure drawdown analysis — where the constants 2.149 and 162.6 come from and how the natural logarithm differs from the decimal one.
Pore pressure and overburden pressure — the hydrostatic constants 0.00981 and 0.052.
Volumetric estimation of hydrocarbons in place — conversion factors for oil and gas in place: acre-feet, barrels, tonnes.

Check yourself

1
The solution gas–oil ratio of an oil is 120 m³/m³, with the gas referred to 20 °C and 101.325 kPa. What is it in scf/STB at 60 °F and 14.696 psia?
1. A.674 scf/STB
2. B.663 scf/STB
3. C.21.4 scf/STB
4. D.120 scf/STB
Show the answer
Correct answer: B. 663 scf/STB
The cubic metre of gas is referred to 20 °C and must be converted to 60 °F, so the factor is 5.529, not 5.6146: $120\cdot5.529 \approx 663$ scf/STB. We get 674 if we leave out the temperature correction (an error of 1.5%), and 21.4 if we divide instead of multiplying.
2
A separator pressure gauge reads 0.35 MPa. Which pressure should we put into the gas equation of state? By how much would the gas volume calculation be wrong if we used the gauge reading?
Show the answer
The absolute pressure is $0.35 + 0.101 = 0.451$ MPa. The volume of a given amount of gas is inversely proportional to the pressure. With 0.35 MPa the volume would come out $0.451/0.35 = 1.29$ times too large, which is 29% too high. The amount of gas in a known volume goes the other way: it would be 22% too low.
3
What does the notation 15 MMscf/d mean?
1. A.15 mega cubic metres per day
2. B.15 thousand standard cubic feet per day
3. C.15 million standard cubic feet per day
4. D.15 billion cubic feet per day
Show the answer
Correct answer: C. 15 million standard cubic feet per day
M is the Roman thousand, and MM is a thousand thousand, which is a million. 15 MMscf/d is about 0.43 million m³/d at 20 °C. A million standard cubic feet at 60 °F is about 28,750 m³ at 20 °C.
4
The oil density at 20 °C is 870 kg/m³. What is that in degrees API and barrels per tonne? What changes if you forget to refer the density to 60 °F?
Show the answer
At 60 °F the oil is denser: $870\cdot(1 + 8\cdot10^{-4}\cdot4.44) = 873.1$ kg/m³. Then $\gamma_o = 873.1/999.0 = 0.874$ and API $= 141.5/0.874 - 131.5 = 30.4$ . A tonne holds $6.2898/0.8731 = 7.20$ barrels. Without the correction we would get 31.0 °API and 7.23 barrels, which is 0.6 of a degree and 0.03 of a barrel too much. The density would be 0.36% too low.
5
The constant 162.6 in the formula for the slope of the semilog straight line is:
1. A.a unit conversion multiplied by ln 10: the formula is written in oilfield units with the decimal logarithm
2. B.a factor for converting psi to MPa
3. C.a property of the reservoir that is found from the test
4. D.a correction for the skin factor
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Correct answer: A. a unit conversion multiplied by ln 10: the formula is written in oilfield units with the decimal logarithm
$162.6 = 70.6\cdot\ln 10$ , and $70.6 = 1/(4\pi\cdot1.127\cdot10^{-3})$ . So the constant contains only unit conversions and the change to the decimal logarithm (where the constants come from). In metric oilfield units the corresponding constant is 2.149.
6
A student wrote the inflow to a well as $q = 2\pi kh\Delta p/(\mu\,r_e/r_w)$ . Will the formula pass the dimensional check? Is it correct?
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It passes: $r_e/r_w$ has no dimension, just as $\ln(r_e/r_w)$ has none. But the formula is wrong. At $r_e/r_w = 2750$ its denominator is $2750/7.92 \approx 350$ times larger than in the Dupuit equation. Dimensions catch errors in units, but not errors in the physics. Deriving the formula and making a check calculation on a known case protect us against such errors.

Updated: 16 October 2026